Câu hỏi của Mun SiNo

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๖ۣۜDũ๖ۣۜN๖ۣۜG · Accepted Answer

7) AQuy đổi hh thành kim loại A (hóa trị n)$A\rightarrow A\left(NO_3\right)_n\rightarrow A_2O_n$$n_{NO_3}=\dfrac{\left(m+62\right)-m}{62}=1\left(mol\right)$=> $n_{A\left(NO_3\right)_n}=\dfrac{1}{n}\left(mol\right)$=> $n_{A_2O_n}=\dfrac{1}{2n}\left(mol\right)$Ta có: $n_A=n_{A\left(NO_3\right)_n}=\dfrac{1}{n}\left(mol\right)$ => $M_A=\dfrac{m}{\dfrac{1}{n}}=m.n\left(g/mol\right)$Ta có: $m_{A_2O_n}=\dfrac{1}{2n}\left(2.M_A+16n\right)=\dfrac{mn}{n}+8=m+8\left(g\right)$8) B$n_{H_2SO_4}=n_{SO_4}=\dfrac{86,6-29}{96}=0,6\left(mol\right)$$\Rightarrow n_{H_2}=0,6\left(mol\right)\Rightarrow V=0,6.22,4=13,44\left(l\right)$9) D$n_{H_2SO_4}=0,1.0,5=0,05\left(mol\right)\Rightarrow n_{H_2O}=0,05\left(mol\right)$Theo ĐLBTKL: moxit + $m_{H_2SO_4}=m_{muối}+m_{H_2O}$=> mmuối = 2,81 + 0,05.98 - 0,05.18 = 6,81 (g)10) C$Ca+2HCl\rightarrow CaCl_2+H_2$$MgO+2HCl\rightarrow MgCl_2+H_2O$Theo PTHH: $n_{HCl}=2.n_{Ca}+2.n_{MgO}=2\left(\dfrac{m_{Ca}}{40}+\dfrac{m_{MgO}}{40}\right)=2.\dfrac{20}{40}=1\left(mol\right)$=> $V=\dfrac{1}{2}=0,5\left(l\right)=500\left(ml\right)$11) APTHH: $Mg+H_2SO_4\rightarrow MgSO_4+H_2$ (1)$Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O$ (2)$MgSO_4+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+Na_2SO_4$ (3) $Fe_2\left(SO_4\right)_3+6NaOH\rightarrow2Fe\left(OH\right)_3\downarrow+3Na_2SO_4$ (4) $Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O$ (5) $2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O$ (6) Gọi số mol Mg, Fe2O3 là a, b (mol)=> 24a + 160b = 20 (1)Theo PTHH (1), (3), (5): $n_{MgO}=a\left(mol\right)$Theo PTHH (2), (4), (6): $n_{Fe_2O_3}=b\left(mol\right)$=> 40a + 160b = 28 (2)(1)(2) => a = 0,5 (mol); b = 0,05 (mol)Theo PTHH (1): $n_{H_2}=0,5\left(mol\right)$=> V = 0,5.22,4 = 11,2 (l)12) A$n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\Rightarrow n_{SO_2}=0,1\left(mol\right)\Rightarrow n_{O_2}=0,05\left(mol\right)$=> $V_{O_2}=0,05.22,4=1,12\left(l\right)$13) B$n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\Rightarrow n_{H_2SO_4}=0,3\left(mol\right)$Theo ĐLBTKL: $m_{KL}+m_{H_2SO_4}=m_{muối}+m_{H_2}$=> mmuối = 14,5 + 0,3.98 - 0,3.2 = 43,3 (g)Câu trả lời: 7) AQuy đổi hh thành kim loại A (hóa trị n)$A\rightarrow A\left(NO_3\right)_n\rightarrow A_2O_n$$n_{NO_3}=\dfrac{\left(m+62\right)-m}{62}=1\left(mol\right)$=> $n_{A\left(NO_3\right)_n}=\dfrac{1}{n}\left(mol\right)$=> $n_{A_2O_n}=\dfrac{1}{2n}\left(mol\right)$Ta có: $n_A=n_{A\left(NO_3\right)_n}=\dfrac{1}{n}\left(mol\right)$ => $M_A=\dfrac{m}{\dfrac{1}{n}}=m.n\left(g/mol\right)$Ta có: $m_{A_2O_n}=\dfrac{1}{2n}\left(2.M_A+16n\right)=\dfrac{mn}{n}+8=m+8\left(g\right)$8) B$n_{H_2SO_4}=n_{SO_4}=\dfrac{86,6-29}{96}=0,6\left(mol\right)$$\Rightarrow n_{H_2}=0,6\left(mol\right)\Rightarrow V=0,6.22,4=13,44\left(l\right)$9) D$n_{H_2SO_4}=0,1.0,5=0,05\left(mol\right)\Rightarrow n_{H_2O}=0,05\left(mol\right)$Theo ĐLBTKL: moxit + $m_{H_2SO_4}=m_{muối}+m_{H_2O}$=> mmuối = 2,81 + 0,05.98 - 0,05.18 = 6,81 (g)10) C$Ca+2HCl\rightarrow CaCl_2+H_2$$MgO+2HCl\rightarrow MgCl_2+H_2O$Theo PTHH: $n_{HCl}=2.n_{Ca}+2.n_{MgO}=2\left(\dfrac{m_{Ca}}{40}+\dfrac{m_{MgO}}{40}\right)=2.\dfrac{20}{40}=1\left(mol\right)$=> $V=\dfrac{1}{2}=0,5\left(l\right)=500\left(ml\right)$11) APTHH: $Mg+H_2SO_4\rightarrow MgSO_4+H_2$ (1)$Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O$ (2)$MgSO_4+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+Na_2SO_4$ (3) $Fe_2\left(SO_4\right)_3+6NaOH\rightarrow2Fe\left(OH\right)_3\downarrow+3Na_2SO_4$ (4) $Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O$ (5) $2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O$ (6) Gọi số mol Mg, Fe2O3 là a, b (mol)=> 24a + 160b = 20 (1)Theo PTHH (1), (3), (5): $n_{MgO}=a\left(mol\right)$Theo PTHH (2), (4), (6): $n_{Fe_2O_3}=b\left(mol\right)$=> 40a + 160b = 28 (2)(1)(2) => a = 0,5 (mol); b = 0,05 (mol)Theo PTHH (1): $n_{H_2}=0,5\left(mol\right)$=> V = 0,5.22,4 = 11,2 (l)12) A$n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\Rightarrow n_{SO_2}=0,1\left(mol\right)\Rightarrow n_{O_2}=0,05\left(mol\right)$=> $V_{O_2}=0,05.22,4=1,12\left(l\right)$13) B$n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\Rightarrow n_{H_2SO_4}=0,3\left(mol\right)$Theo ĐLBTKL: $m_{KL}+m_{H_2SO_4}=m_{muối}+m_{H_2}$=> mmuối = 14,5 + 0,3.98 - 0,3.2 = 43,3 (g)

Kudo Shinichi · Answer

7)Áp dụng ĐLBTNT: $m_{-NO_3}=\left(m+62\right)-m=62\left(g\right)$$\rightarrow n_{-NO_3}=\dfrac{62}{62}=1\left(mol\right)\
\xrightarrow[]{BTNT.N}n_{NO_2}=1\left(mol\right)$PTHH:$4Fe\left(NO_3\right)_3\xrightarrow[t^o]{}2Fe_2O_3+12NO_2+3O_2$$2Zn\left(NO_3\right)_2\xrightarrow[]{t^o}2ZnO+4NO_2+O_2\
2Cu\left(NO_3\right)_2\xrightarrow[]{t^o}2CuO+4NO_2+O_2$Theo PTHH: $n_{O_2}=\dfrac{1}{4}n_{NO_2}=0,25\left(mol\right)$Áp dụng ĐLBTKL:$m_{oxit}=\left(m+62\right)-0,25.32-46=\left(m+8\right)\left(g\right)$=> Chọn A8)Áp dụng ĐLBTNT: $n_{=SO_4}=86,6-29=57,6\left(g\right)$$\rightarrow n_{=SO_4}=\dfrac{57,6}{96}=0,6\left(mol\right)$PTHH:$Mg+H_2SO_4\rightarrow MgSO_4+H_2$$Zn+H_2SO_4\rightarrow ZnSO_4+H_2$$Fe+H_2SO_4\rightarrow FeSO_4+H_2$Theo PTHH: $n_{H_2}=n_{=SO_4}=0,6\left(mol\right)$=> V = 0,6.22,4 = 13,44 (l)=> Chọn B9)$n_{H_2SO_4}=0,5.0,1=0,05\left(mol\right)\
\rightarrow n_{=SO_4}=0,05\left(mol\right)$PTHH:$Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O$$ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O$$MgO+H_2SO_4\rightarrow MgSO_4+H_2O$Theo PTHH: $n_{O\left(oxit\right)}=n_{=SO_4}=0,05\left(mol\right)$Theo ĐLBTNT: $m_{muối}=m_{oxit}-m_O+m_{=SO_4}=2,81-0,05.16+0,05.96=6,81\left(g\right)$=> Chọn D10)Ta có: $M_{Ca}=M_{MgO}$$\rightarrow\overline{M_{hh}}=40\left(g\text{/}mol\right)\
\rightarrow n_{hh}=\dfrac{20}{40}=0,5\left(mol\right)$PTHH:$Ca+2HCl\rightarrow CaCl_2+H_2\
MgO+2HCl\rightarrow MgCl_2+H_2O$Theo PTHH: nHCl = 2nhh = 1 (mol)=> $V_{dd.HCl}=\dfrac{1}{2}=0,5\left(l\right)=500\left(ml\right)$=> Chọn C11)$m_{hh.sau.các.pư}=m_{O\left(MgO\right)}=8\left(g\right)\
\rightarrow n_O=\dfrac{8}{16}=0,5\left(mol\right)\
\rightarrow n_{Mg}=n_O=0,5\left(mol\right)$PTHH: Mg + H2SO4 ---> MgSO4 + H2           0,5------------------------------->0,5=> VH2 = 0,5.22,4 = 11,2 (l)=> Chọn A12)$n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)$Theo PTHH: $n_{O_2}=\dfrac{1}{2}n_{SO_2}=\dfrac{1}{2}n_{Cu}=0,05\left(mol\right)$=> VO2 = 0,05.22,4 = 1,12 (l)=> Chọn A13)PTHH:$Mg+H_2SO_4\rightarrow MgSO_4+H_2$$Zn+H_2SO_4\rightarrow ZnSO_4+H_2$$Fe+H_2SO_4\rightarrow FeSO_4+H_2$$n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)$Theo PTHH: $n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)$Áp dụng ĐLBTKL:$m_{muối}=14,5+0,3.98-0,3.2=43,3\left(g\right)$=> Chọn BCâu trả lời: 7)Áp dụng ĐLBTNT: $m_{-NO_3}=\left(m+62\right)-m=62\left(g\right)$$\rightarrow n_{-NO_3}=\dfrac{62}{62}=1\left(mol\right)\
\xrightarrow[]{BTNT.N}n_{NO_2}=1\left(mol\right)$PTHH:$4Fe\left(NO_3\right)_3\xrightarrow[t^o]{}2Fe_2O_3+12NO_2+3O_2$$2Zn\left(NO_3\right)_2\xrightarrow[]{t^o}2ZnO+4NO_2+O_2\
2Cu\left(NO_3\right)_2\xrightarrow[]{t^o}2CuO+4NO_2+O_2$Theo PTHH: $n_{O_2}=\dfrac{1}{4}n_{NO_2}=0,25\left(mol\right)$Áp dụng ĐLBTKL:$m_{oxit}=\left(m+62\right)-0,25.32-46=\left(m+8\right)\left(g\right)$=> Chọn A8)Áp dụng ĐLBTNT: $n_{=SO_4}=86,6-29=57,6\left(g\right)$$\rightarrow n_{=SO_4}=\dfrac{57,6}{96}=0,6\left(mol\right)$PTHH:$Mg+H_2SO_4\rightarrow MgSO_4+H_2$$Zn+H_2SO_4\rightarrow ZnSO_4+H_2$$Fe+H_2SO_4\rightarrow FeSO_4+H_2$Theo PTHH: $n_{H_2}=n_{=SO_4}=0,6\left(mol\right)$=> V = 0,6.22,4 = 13,44 (l)=> Chọn B9)$n_{H_2SO_4}=0,5.0,1=0,05\left(mol\right)\
\rightarrow n_{=SO_4}=0,05\left(mol\right)$PTHH:$Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O$$ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O$$MgO+H_2SO_4\rightarrow MgSO_4+H_2O$Theo PTHH: $n_{O\left(oxit\right)}=n_{=SO_4}=0,05\left(mol\right)$Theo ĐLBTNT: $m_{muối}=m_{oxit}-m_O+m_{=SO_4}=2,81-0,05.16+0,05.96=6,81\left(g\right)$=> Chọn D10)Ta có: $M_{Ca}=M_{MgO}$$\rightarrow\overline{M_{hh}}=40\left(g\text{/}mol\right)\
\rightarrow n_{hh}=\dfrac{20}{40}=0,5\left(mol\right)$PTHH:$Ca+2HCl\rightarrow CaCl_2+H_2\
MgO+2HCl\rightarrow MgCl_2+H_2O$Theo PTHH: nHCl = 2nhh = 1 (mol)=> $V_{dd.HCl}=\dfrac{1}{2}=0,5\left(l\right)=500\left(ml\right)$=> Chọn C11)$m_{hh.sau.các.pư}=m_{O\left(MgO\right)}=8\left(g\right)\
\rightarrow n_O=\dfrac{8}{16}=0,5\left(mol\right)\
\rightarrow n_{Mg}=n_O=0,5\left(mol\right)$PTHH: Mg + H2SO4 ---> MgSO4 + H2           0,5------------------------------->0,5=> VH2 = 0,5.22,4 = 11,2 (l)=> Chọn A12)$n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)$Theo PTHH: $n_{O_2}=\dfrac{1}{2}n_{SO_2}=\dfrac{1}{2}n_{Cu}=0,05\left(mol\right)$=> VO2 = 0,05.22,4 = 1,12 (l)=> Chọn A13)PTHH:$Mg+H_2SO_4\rightarrow MgSO_4+H_2$$Zn+H_2SO_4\rightarrow ZnSO_4+H_2$$Fe+H_2SO_4\rightarrow FeSO_4+H_2$$n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)$Theo PTHH: $n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)$Áp dụng ĐLBTKL:$m_{muối}=14,5+0,3.98-0,3.2=43,3\left(g\right)$=> Chọn B

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